   Chapter 18, Problem 91SCQ

Chapter
Section
Textbook Problem

Write a chemical equation for the oxidation of C2H6(g) by O2(g) to form CO2(g) and H2O(g). Defining this as the system: (a) Predict whether the signs of ΔS°(system), ΔS° (surroundings), and ΔS°(universe) will be greater than zero, equal to zero, or less than zero. Explain your prediction. (b) Predict the signs of ΔrH° and ΔrG°. Explain how you made this prediction. (c) Will the value of Kp be very large, very small, or near 1? Will the equilibrium constant, Kp, for this system be larger or smaller at temperatures greater than 298 K? Explain how you made this prediction.

(a)

Interpretation Introduction

Interpretation:

It should be explained that whether the sign of entropy change for system, surroundings and universe is greater than zero or not.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

ΔSo(universe)= ΔSo(system)+ΔSo(surroundings)

The ΔSo(universe) should be greater than zero for a spontaneous process.

The  ΔSo(system) can be calculated by the following expression,

ΔSo(system)= ΔrS°nS°(products)-nS°(reactants)

The ΔSo(surroundings) can be calculated by the following expression,

ΔSo(surroundings)=- ΔrHoT

Here, ΔrH° is the enthalpy change for the reaction.

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔGo. It is related to entropy and entropy by the following expression,

ΔrGorHo-TΔrSo

ΔGo is also related to the equilibrium constant K by the equation,

ΔrGo= -RTlnKp

The rearranged expression is,

Kp= eΔrGoRT

Explanation

The oxidation of C2H6(g) by O2(g) forms CO2(g) and H2O(g). The balanced chemical equation is,

2C2H6(g)+7O2(g)4CO2

(b)

Interpretation Introduction

Interpretation:

The sign of ΔrH° and ΔrGo value for the given reaction should be predicted.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

ΔSo(universe)= ΔSo(system)+ΔSo(surroundings)

The ΔSo(universe) should be greater than zero for a spontaneous process.

The  ΔSo(system) can be calculated by the following expression,

ΔSo(system)= ΔrS°nS°(products)-nS°(reactants)

The ΔSo(surroundings) can be calculated by the following expression,

ΔSo(surroundings)=- ΔrHoT

Here, ΔrH° is the enthalpy change for the reaction.

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔGo. It is related to entropy and entropy by the following expression,

ΔrGorHo-TΔrSo

ΔGo is also related to the equilibrium constant K by the equation,

ΔrGo= -RTlnKp

The rearranged expression is,

Kp= eΔrGoRT

(c)

Interpretation Introduction

Interpretation:

It should be identified that whether the value of Kp is very large, very small or nearer to 1 also that the equilibrium constant will be larger or smaller for temperatures greater than 298K.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

ΔSo(universe)= ΔSo(system)+ΔSo(surroundings)

The ΔSo(universe) should be greater than zero for a spontaneous process.

The  ΔSo(system) can be calculated by the following expression,

ΔSo(system)= ΔrS°nS°(products)-nS°(reactants)

The ΔSo(surroundings) can be calculated by the following expression,

ΔSo(surroundings)=- ΔrHoT

Here, ΔrH° is the enthalpy change for the reaction.

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔGo. It is related to entropy and entropy by the following expression,

ΔrGorHo-TΔrSo

ΔGo is also related to the equilibrium constant K by the equation,

ΔrGo= -RTlnKp

The rearranged expression is,

Kp= eΔrGoRT

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